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How to calculate this question? $$\lim\limits_{{\rho}\rightarrow 0^+}\frac{\log{(1-(a^{-\rho}+b^{-\rho}-c^{-\rho}))}}{\log{\rho}} ,$$

where $a>0$, $b>0$ and $c>0$.

It's similarly to another question How to calculate $\lim\limits_{{\rho}\rightarrow 0^+}\frac{\log{(1-(a^{-\rho}+b^{-\rho}-(ab)^{-\rho}))}}{\log{\rho}} $ with $a>1$ and $b>1$?.

Thank you everyone.

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Let's get the answer via series expansions. We have $$a^{-\rho} = \exp(-\rho\log a) = 1 - \rho\log a + o(\rho)$$ and hence $$a^{-\rho} + b^{-\rho} - c^{-\rho} = 1 - \rho\log(ab/c) + o(\rho)$$ so that $$\log(1 - (a^{-\rho} + b^{-\rho} - c^{-\rho})) = \log(\rho\log(ab/c)) + o(\rho))$$ and hence we see that we must have $\log(ab/c) > 0$ for the above expression to be defined. Thus we must have $ab > c$ (the case $ab = c$ has been covered in the question linked in OP's post). And then we can see that $$\log(1 - (a^{-\rho} + b^{-\rho} - c^{-\rho})) = \log \rho + \log\log(ab/c) + o(1)$$ and therefore $$\frac{\log(1 - (a^{-\rho} + b^{-\rho} - c^{-\rho}))}{\log\rho} = 1 + \frac{\log\log(ab/c)}{\log\rho} + o(1)$$ and hence the desired limit is $1$.

Note that the case $ab = c$ can also be handled via series expansions but then $\log(ab/c) = 0$ so the term with $\rho\log(ab/c)$ vanishes and hence we need the series expansions till $\rho^{2}$. But we don't need to calculate the coefficient of this $\rho^{2}$. We just have $$a^{-\rho} + b^{-\rho} - c^{-\rho} = 1 + k\rho^{2} + o(\rho^{2})$$ and then $$\log(1 - (a^{-\rho} + b^{-\rho} - c^{-\rho})) = \log k + 2\log\rho + o(1)$$ and when we divide by $\log \rho$ we can see that the desired limit is $2$.

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  • $\begingroup$ thx, nice answer ! $\endgroup$
    – Leung
    Mar 28 '16 at 6:28
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The limit is equivalent to

$$\lim_{p\to 0}\frac{p(log(a)a^{-p}+log(b)b^{-p}-log(c)c^{-p})}{1-a^{-p}-b^{-p}+c^{-p}}$$

L'Hopital again,

$$\lim_{p\to 0}\frac{(log(a)a^{-p}+log(b)b^{-p}-log(c)c^{-p})+p(-log(a)^2a^{-p}-log(b)^2b^{-p}+log(c)^2c^{-p})}{log(a)a^{-p}+log(b)b^{-p}-log(c)c^{-p}}$$

If $ab\neq c$, then it is just $1$. If the $ab=c$, it is $2$ (same as your previous question)

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    $\begingroup$ There is the tag limits-without-lhopital $\endgroup$ Mar 27 '16 at 8:08
  • $\begingroup$ @ClaudeLeibovici Sorry, I didn't see that. I will try to think of an alternative answer $\endgroup$
    – lEm
    Mar 27 '16 at 8:23

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